Every integer is a square times a square-free integer

Statement

Let $n$ be a positive integer. Then there are positive integers $a$ and $b$ such that $n = a^2 b$ and $b$ is square-free.

Remarks

The integers $a$ and $b$ are unique. The number $b$ is the product of the primes $p$ for which the exponent of $p$ in $n$ is odd. For example, $72 = 2^3 \cdot 3^2 = 6^2 \cdot 2$, so $a = 6$ and $b = 2$. The proof uses the prime factorization of $n$. Write each exponent $e$ as $e = 2 q + r$ with $r \in \{0, 1\}$.

Explanations

L1 human

Take out of $n$ all the pairs of equal primes that you can. The pairs give the square $a^2$. At most one copy of each prime stays, so the rest $b$ is square-free. For example, $50 = 2 \cdot 5 \cdot 5 = 5^2 \cdot 2$.